| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 2363 | Accepted: 881 |
Description
Like so many others, the cows have developed very haughty tastes and will no longer graze on just any grass. Instead, Farmer John must purchase gourmet organic grass at the Green Grass Grocers store for each of his N (1 ≤ N ≤ 100,000) cows.
Each cow i demands grass of price at least Ai (1 ≤ Ai ≤ 1,000,000,000) and with a greenness score at least Bi (1 ≤ Bi ≤ 1,000,000,000). The GGG store has M (1 ≤ M ≤ 100,000) different types of grass available, each with a price Ci (1 ≤ Ci ≤ 1,000,000,000) and a greenness score of Di (1 ≤ Di ≤ 1,000,000,000). Of course, no cow would sacrifice her individuality, so no two cows can have the same kind of grass.
Help Farmer John satisfy the cows' expensive gourmet tastes while spending as little money as is necessary.
Input
* Line 1: Two space-separated integers: N and M.
* Lines 2..N+1: Line i+1 contains two space-separated integers: Ai and Bi
* Lines N+2..N+M+1: Line i+N+1 contains two space-separated integers: Ci and Di
Output
* Line 1: A single integer which is the minimum cost to satisfy all the cows. If that is not possible, output -1.
Sample Input
4 7 1 1 2 3 1 4 4 2 3 2 2 1 4 3 5 2 5 4 2 6 4 4
Sample Output
12
Source
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<ctime> 5 #include<cstdlib> 6 #include<algorithm> 7 using namespace std; 8 #define ls T[now].ch[0] 9 #define rs T[now].ch[1] 10 const int MAXN=3*1e5+10; 11 inline char nc() 12 { 13 static char buf[MAXN],*p1=buf,*p2=buf; 14 return p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++; 15 } 16 inline int read() 17 { 18 char c=nc();int x=0,f=1; 19 while(c<'0'||c>'9'){if(c=='-')f=-1;c=nc();} 20 while(c>='0'&&c<='9'){x=x*10+c-'0',c=nc();} 21 return x*f; 22 } 23 struct node 24 { 25 int val,ch[2],pri,siz; 26 }T[MAXN]; 27 int x,y,z,root=0,pos; 28 int tot=0; 29 void update(int now) 30 { 31 T[now].siz=T[ls].siz+T[rs].siz+1; 32 } 33 inline int newnode(int v) 34 { 35 T[++tot].val=v; 36 T[tot].pri=rand(); 37 T[tot].siz=1; 38 return tot; 39 } 40 int merge(int x,int y) 41 { 42 if(!x||!y) return x+y; 43 if(T[x].pri<T[y].pri) 44 { 45 T[x].ch[1]=merge(T[x].ch[1],y); 46 update(x); 47 return x; 48 } 49 else 50 { 51 T[y].ch[0]=merge(x,T[y].ch[0]); 52 update(y); 53 return y; 54 } 55 } 56 void split(int now,int k,int &x,int &y) 57 { 58 if(!now) {x=y=0;return ;} 59 if(T[now].val<=k) x=now,split(rs,k,rs,y); 60 else y=now,split(ls,k,x,ls); 61 update(now); 62 } 63 struct N 64 { 65 int x,y; 66 }a[MAXN],b[MAXN]; 67 int comp(const N &a,const N &b) 68 { 69 return a.x<b.x||(a.x==b.x&&a.y<b.y); 70 } 71 void insert(int val) 72 { 73 split(root,val,x,y); 74 root=merge( merge(x,newnode(val)) , y ); 75 } 76 int kth(int now,int x) 77 { 78 while(now) 79 { 80 if(T[ls].siz+1==x) return now; 81 else if(T[ls].siz>=x) now=ls; 82 else x-=T[ls].siz+1,now=rs; 83 } 84 } 85 void Delet(int now) 86 { 87 split(root,now,x,z); 88 split(x,now-1,x,y); 89 y=merge(T[y].ch[0] , T[y].ch[1] ); 90 root=merge( merge(x,y) ,z ); 91 } 92 int main() 93 { 94 #ifdef WIN32 95 freopen("a.in","r",stdin); 96 #else 97 #endif 98 int n,m; 99 scanf("%d%d",&n,&m); 100 if (m<n){puts("-1");return 0;} 101 for(int i=1;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].y); 102 for(int i=1;i<=m;i++) scanf("%d%d",&b[i].x,&b[i].y); 103 sort(a+1,a+n+1,comp); 104 sort(b+1,b+m+1,comp); 105 long long int cur=1,ans=0,num=0;//在第一个中找到了几个 106 for(int i=1;i<=m;i++) 107 { 108 while(a[cur].x<=b[i].x&&cur<=n) 109 insert(a[cur].y),cur++; 110 split(root,b[i].y,x,y); 111 pos=kth(x,T[x].siz); 112 if(pos==0) continue; 113 num++; 114 root=merge(x,y); 115 Delet(T[pos].val); 116 ans+=b[i].x; 117 } 118 if(num==n) printf("%lld",ans); 119 else printf("-1"); 120 return 0; 121 }